# Conservation of Momentum of a spring

1. Nov 10, 2009

### interxavier

1. The problem statement, all variables and given/known data
A hollow box, mass M, rests on a frictionless surface. Inside the box is a spring, spring constant k, attached to one wall. A block, mass m, is pushed against the spring so that it is compressed an amount A.
a. If after leaving the spring the block has velocity of magnitude v1, what will be the velocity of the box?
b. The block slides across and hits the opposite side of the box. Assuming this collision to be perfectly elastic, find the velocities of the box and te block after the collision.
c. What is v1?

2. Relevant equations
p = mv
m1*v1 + m2*v2 = m1*u1 + m2*u2

"u" is final velocity.

3. The attempt at a solution

I'm lost. I know the spring is gonna exert a force on the box, so the box moves in the opposite direction. That's all I know.

2. Nov 10, 2009

### Tzim

I dont know which one is correct so i will tell you two cases.(if some of them is correct)
1)I think that until the body leaves the spring we donnto have a conservation of momentum because the system Box-body(m) is not isolated as rhe force from spring ecxerts.When though the ball leaves the spring there is a conservation of momenum and then you can take the formula for the Box(M)-body(m).

2)If you take as your system Box-spring-body(m)
then the spring excert a force on the body and an opossite and equal force to the Box.So i theoritically believe that the velocity equation for the Box should be something like U=Umax*cos(ωt+φο) where ω^2=K/M.

The second case is very unlike to happen because the spring is fixed on the wall of the box so plz try the fisrt case and let me know if i helped you or if i have written anything very stupid.

3. Nov 10, 2009

### Tzim

In both cases if you find the final speed of the mass(m) then you just use conservation of momentum

4. Nov 10, 2009

### Vaal

a) you have the formula right you just need to plug in the correct values and solve. We can say:
m1=m, the mass of the block
m2=M, the mass of the box
v2=0, again they are not moving to start
u1=v1, the problem says the block's velocity is v1 after being pushed by the spring (be sure not to confuse this with v1 above)
u2 is the velocity of box that you are solving for

b) On this part you must use your conservation of momentum equation from "a" and the conservation of energy equation,
.5m1v12 + .5m2v22 = .5m1u12 + .5m2u22, since the collision is perfectly elastic. You will use the final velocities from problem "a" (u1 and u2) as your initial velocities(v1 and v2) in problem "b" as problem "b" is happening after problem "a". This should give you a system of equations that you can solve to obtain your new final velocities.

c) Here I assume they are talking about v1 from "a". If this is the case the easiest thing to do is use conservation of energy.
The energy imparted on the block by the spring is given by:
.5kx2 where,
k= spring constant
x=compression distance
all the energy will be kinetic so set this equal to the formula for the kinetic energy of the block, .5mv2, and you will be able to solve for velocity.

Hope this helps.